# Direction of motion climbing up stairs for work problem

I wanted to know why the we calculate the work done of a person climbing up a set of stairs using the y direction,for example, we use the equation w=mgh to calculate the work done for a person climbing up a set of stairs. I do not understand why we are not also calculating the work done in x direction because the direction of the motion on a set of stairs is parrallel to the stairs so my I would think that the total work would be the hypotenuse of a work triangle not the y component by itself.

Chestermiller
Mentor
Suppose you're sliding a chunk of ice of mass M up an inclined plane (plane angle θ). There is no friction between the plane and the ice. The distance you push the ice along the slide is L, but the vertical distance that the ice moves is h=Lsinθ. What force do you need to apply to the ice along the incline? How much work do you do?

Chet

Suppose you're sliding a chunk of ice of mass M up an inclined plane (plane angle θ). There is no friction between the plane and the ice. The distance you push the ice along the slide is L, but the vertical distance that the ice moves is h=Lsinθ. What force do you need to apply to the ice along the incline? How much work do you do?

Chet
nonetheless the direction of the motion is parrellel to the angle of the inclined planed, the motion has to components of distance, I am sorry but your reply is concise but does not answer my question. I am thinking that the work is determined by where the direction of the force and motion are together, maybe it is the direction of the momtion that has a force acting on the dimension?

Chestermiller
Mentor
nonetheless the direction of the motion is parrellel to the angle of the inclined planed, the motion has to components of distance, I am sorry but your reply is concise but does not answer my question. I am thinking that the work is determined by where the direction of the force and motion are together, maybe it is the direction of the momtion that has a force acting on the dimension?
The force required to push the ice cube up the ramp is mgsinθ. It is pushed through a distance L along the ramp. So the work is mgsinθL. But, L=h/sinθ. So the work is also mgh, where h is the vertical distance. Do you see what I'm driving at, and how this relates to your staircase question?

Chet

A.T.
I do not understand why we are not also calculating the work done in x direction because the direction of the motion
Because we ignore the horizontal forces, and consider only the vertical ones. For constant speed and slope the horizontal work averages to zero, but the body still dissipates energy to cyclically accelerate horizontally.

PeroK
Homework Helper
Gold Member
2021 Award
I wanted to know why the we calculate the work done of a person climbing up a set of stairs using the y direction,for example, we use the equation w=mgh to calculate the work done for a person climbing up a set of stairs. I do not understand why we are not also calculating the work done in x direction because the direction of the motion on a set of stairs is parrallel to the stairs so my I would think that the total work would be the hypotenuse of a work triangle not the y component by itself.

If you walk along a flat piece of ground, how much work do you do? Or, better, if you're cycling and you stop peddling, how much work do you do to keep the bike moving?

Because you are treating the motion according to a simplified model. Many people say that you do no work while walking, but this is not true. Your center of mass rises and falls as you walk, but what force is doing work? Is it friction? No. Static friction cannot do work, it can only transmit force. The force(s) responsible for the work done on your center are due to the movement of your legs. A similar question that often confuses people is to ask what force is responsible for doing work on a car as it increases speed on a flat road. Again, friction is not doing the work; it is only transmitting the force.

• Jewish_Vulcan
You are not giving me any clear scientific answers only examples! I have an idea, since work is a dot vector product then the work can only be done where the motion and the force are in the same direction? Furthermore why is there no work while walking on a flat surface?

If you walk along a flat piece of ground, how much work do you do? Or, better, if you're cycling and you stop peddling, how much work do you do to keep the bike moving?
I do not know stop with the theory questions and give me a direct answer.... My teacher cannot even explain it.

Because we ignore the horizontal forces, and consider only the vertical ones. For constant speed and slope the horizontal work averages to zero, but the body still dissipates energy to cyclically accelerate horizontally.
My question is why we ignore the horizontal forces. How does it average to 0, I want to know why I am thinking wrong about there being work in the x direction climbing up an inclined plane.

Correct. Work is only done by a force when it is in the same direction as the motion.

I never said that there was no net horizontal work being done. That is a bold assumption. In fact, check out this article.

The only reason you are ignoring the horizontal work is to simplify the calculations. You are making an approximation. If you do not make this approximation you would need to model the mechanics of walking. I bet that your teacher would be impressed if you included this in your calculations!

• Jewish_Vulcan
Correct. Work is only done by a force when it is in the same direction as the motion.

I never said that there was no net horizontal work being done. That is a bold assumption. In fact, check out this article.

The only reason you are ignoring the horizontal work is to simplify the calculations. You are making an approximation. If you do not make this approximation you would need to model the mechanics of walking. I bet that your teacher would be impressed if you included this in your calculations!
Well that is all I wanted to know, try to understand that it was frusterating when people were telling me that there was no work in the horizontal direction. I did not want to know how but why, It made sense that there should be some more complex forces acting horizontally when walking, these people were mostly giving me examples like "suppose"! Thank you for giving me a direct answer.

A.T.
My question is why we ignore the horizontal forces. How does it average to 0
Since we don't accumulate horizontal velocity, the positive and negative work done by horizontal forces balance each other over the course of a gait cycle.

CWatters
Homework Helper
Gold Member
Since we don't accumulate horizontal velocity, the positive and negative work done by horizontal forces balance each other over the course of a gait cycle.

That's true but it's not really why it's ignored. You could say the same about the vertical velocity.

Well that is all I wanted to know, try to understand that it was frusterating when people were telling me that there was no work in the horizontal direction. I did not want to know how but why, It made sense that there should be some more complex forces acting horizontally when walking, these people were mostly giving me examples like "suppose"! Thank you for giving me a direct answer.

You are likely to come across a lot of problems that require you to decide what factors to ignore or simplify. Usually if a force isn't specified in the problem you can't take it into account. Beware problems that contain data you don't need.

Humans need energy just to live so you actually made your own simplification by ignoring energy consumed breathing, friction in joints etc.

A.T.
You could say the same about the vertical velocity.
Yes, in this type of analysis we also ignore the work done by vertical forces to accelerate the body. We consider just the m*g part of the vertical force.

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I think part of the confusion with this question is it was stated as 'work done by a person,' but a person is not a force. To avoid confusion when discussing work I prefer to phrase things as work done by ___ (insert force here) on _____ (insert object here).

This is why I posted the article which provides evidence that the amount of positive work is done by the muscles in the human body is greater than the amount of negative work done by the muscles. So when considering the biomechanics of walking the 'person' does a nonzero amount of work in the horizontal direction.

• Jewish_Vulcan
jbriggs444
Homework Helper
I think part of the confusion with this question is it was stated as 'work done by a person,' but a person is not a force. To avoid confusion when discussing work I prefer to phrase things as work done by ___ (insert force here) on _____ (insert object here).

This is why I posted the article which provides evidence that the amount of positive work is done by the muscles in the human body is greater than the amount of negative work done by the muscles. So when considering the biomechanics of walking the 'person' does a nonzero amount of work in the horizontal direction.

That article says nothing about the amount of "work in the horizontal direction" that was done. It merely says that the work done by the muscles is positive in spite of the fact that neither kinetic nor potential energy of the body increases as a result. But that is unremarkable. Any man who has watched a female jogger on a level road knows where at least a portion of the work done by her muscles is going.

In the words of the article: "elicited greater energy dissipation in non-muscular tissues"

You're absolutely right. My apologies.

• Jewish_Vulcan
sophiecentaur
Gold Member
@brianpushups
With a monicker like that, you should know all about doing a lot of work and not actually going anywhere. ;)

Ha! Boy do I!

Am I correct in concluding that overall there is no work in the horizontal direction because there is no force in the horizontal direction although there is motion in the horizontal direction? I thought that for something to be in motion it required a force..

Force is not required to maintain motion; that is exactly what Newton's first law states (in the absence of a net force an object will move with constant velocity). Suppose that something is in motion along the horizontal direction at 3m/s. If a 5N force pulls to the right and a 5N force pulls to the left the net force is zero and the object will still move at 3m/s.

Similarly, the net work requires a net force. If there is no net horizontal force there will be no net horizontal work.

• Jewish_Vulcan
Force is not required to maintain motion; that is exactly what Newton's first law states (in the absence of a net force an object will move with constant velocity). Suppose that something is in motion along the horizontal direction at 3m/s. If a 5N force pulls to the right and a 5N force pulls to the left the net force is zero and the object will still move at 3m/s.

Similarly, the net work requires a net force. If there is no net horizontal force there will be no net horizontal work.
My intuition tells me that if there is a net force of 0 or a balanced force then nothing would change in the system. It would only still be going 3m/s if it had an intitial unbalanced force that caused it to go 3m/s. I was always considering the initial force in the horizontal direction when I was thinking of work... that is why I thought there should be work in the horizontal direction. OVERALL BECAUSE THE FORCE IN THE HORIZONTAL DIRECTION IS NOT CONSTANT THERE IS NOT WORK IN THE HORIZONTAL DIRECTION.Thank you brainpushups, I think I understand now.

Cool. You are right: If you are initially at rest then work must be done to get you moving - force changes the state of rest and that force acts over a distance. Once the walking speed is achieved there is both positive and negative work being done by your muscles on your center of mass. To stop walking your muscles must do negative work in the horizontal direction.

Cool. You are right: If you are initially at rest then work must be done to get you moving - force changes the state of rest and that force acts over a distance. Once the walking speed is achieved there is both positive and negative work being done by your muscles on your center of mass. To stop walking your muscles must do negative work in the horizontal direction.
Would not there be work in the horizontal direction initially?

Chestermiller
Mentor
Would not there be work in the horizontal direction initially?
Yes. The step would exert a horizontal force on you that would create the horizontal kinetic energy (3m/s velocity in your example), but when you stopped at the top of the stairs, equal work would be done in the opposite direction to stop you. If you add the two amounts of work together, it sums to zero.

Chet

• Jewish_Vulcan
Yes. The step would exert a horizontal force on you that would create the horizontal kinetic energy (3m/s velocity in your example), but when you stopped at the top of the stairs, equal work would be done in the opposite direction to stop you. If you add the two amounts of work together, it sums to zero.

Chet
That means that we are assuming that velocity of the object goes to 0 after the distance, maybe they should have said that it traveled that distance and then came to a stop! Thanks for helping me understand why there is no net work in the horizontal direction chet.

The work does not have a direction so it does not make sense to say "there is no work in the horizontal direction".
It may be part of your initial confusion too, as well as the ambiguous formulation of the problem in OP.
"mgh" is not the "work done for a person climbing up a set of stairs" but is the (absolute value of) the work done by the force of gravity on that person. As gravity is always vertical, only the vertical displacement count for the work of this force.
There are other forces acting on the person and the work of these forces can have any values, depending on the specific problem. It is not necessary for the work of these other forces to be zero or have any specific value.

However, if the initial and final kinetic energies are the same (maybe zero) then the work of the net force will be zero (the work-energy theorem). Because gravity is doing -mgh, the work of all the other forces will add up to +mgh.

Nasu, while you are certainly correct in saying that work does not have a direction I don't think this was contributing to the confusion. Once a coordinate system is chosen the total work can be viewed the sum total of the work along each of three orthogonal directions. If there is a horizontal component of displacement and a horizontal component of force then I think we can say that some work was done along the horizontal direction

• Jewish_Vulcan
A.T.
maybe they should have said that it traveled that distance and then came to a stop!
Why? Did they specify that it started from rest? If nothing is stated about the initial vs. final velocity you should naturally assume they are equal, but not necessarily zero.

The work does not have a direction so it does not make sense to say "there is no work in the horizontal direction".
It may be part of your initial confusion too, as well as the ambiguous formulation of the problem in OP.
"mgh" is not the "work done for a person climbing up a set of stairs" but is the (absolute value of) the work done by the force of gravity on that person. As gravity is always vertical, only the vertical displacement count for the work of this force.
There are other forces acting on the person and the work of these forces can have any values, depending on the specific problem. It is not necessary for the work of these other forces to be zero or have any specific value.

However, if the initial and final kinetic energies are the same (maybe zero) then the work of the net force will be zero (the work-energy theorem). Because gravity is doing -mgh, the work of all the other forces will add up to +mgh.
whatever... that response did not even help at all.
Why? Did they specify that it started from rest? If nothing is stated about the initial vs. final velocity you should naturally assume they are equal, but not necessarily zero.
I do not "assume facts", I am a very literal person.If they are more clear on the parameters of the problem I would have understood why there is a net force of 0 in the horizontal direction. That is like saying assuming that the velocity initial is 0 without them stating it!

A.T.